9.2 - More Examples

Let's take a look at two more examples of a hypothesis test for a single proportion while recalling the hypothesis testing procedure we outlined on the previous page:

  1. State the null hypothesis \(H_0\) and the alternative hypothesis \(H_{A}\).

  2. Calculate the test statistic:

    \(Z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}\)

  3. Determine the critical region.

  4. Make a decision. Determine if the test statistic falls in the critical region. If it does, reject the null hypothesis. If it does not, do not reject the null hypothesis.

The first example involves a hypothesis test for the proportion in which the alternative hypothesis is a "greater than hypothesis," that is, the alternative hypothesis is of the form \(H_A \colon p > p_0\). And, the second example involves a hypothesis test for the proportion in which the alternative hypothesis is a "less than hypothesis," that is, the alternative hypothesis is of the form \(H_A \colon p < p_0\).

Example 9-2 Section

seat belt

Let p equal the proportion of drivers who use a seat belt in a state that does not have a mandatory seat belt law. It was claimed that \(p = 0.14\). An advertising campaign was conducted to increase this proportion. Two months after the campaign, \(y = 104\) out of a random sample of \(n = 590\) drivers were wearing seat belts. Was the campaign successful?

Answer

The observed sample proportion is:

\(\hat{p}=\dfrac{104}{590}=0.176\)

Because it is claimed that \(p = 0.14\), the null hypothesis is:

\(H_0 \colon p = 0.14\)

Because we're interested in seeing if the advertising campaign was successful, that is, that a greater proportion of people wear seat belts, the alternative hypothesis is:

\(H_A \colon p > 0.14\)

The test statistic is therefore:

\(Z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.176-0.14}{\sqrt{\dfrac{0.14(0.86)}{590}}}=2.52\)

If we use a significance level of \(\alpha = 0.01\), then the critical region is:

Zα = 0.012.326

That is, we reject the null hypothesis if the test statistic \(Z > 2.326\). Because the test statistic falls in the critical region, that is, because \(Z = 2.52 > 2.326\), we can reject the null hypothesis in favor of the alternative hypothesis. There is sufficient evidence at the \(\alpha = 0.01\) level to conclude the campaign was successful (\(p > 0.14\)).

Again, note that this is an example of a right-tailed hypothesis test because the action falls in the right tail of the normal distribution.

Example 9-3 Section

very happy man

A Gallup poll released on October 13, 2000, found that 47% of the 1052 U.S. adults surveyed classified themselves as "very happy" when given the choices of:

  • "very happy"
  • "fairly happy"
  • "not too happy"

Suppose that a journalist who is a pessimist took advantage of this poll to write a headline titled "Poll finds that U.S. adults who are very happy are in the minority." Is the pessimistic journalist's headline warranted?

Answer

The sample proportion is:

\(\hat{p}=0.47\)

Because we're interested in the majority/minority boundary line, the null hypothesis is:

\(H_0 \colon p = 0.50\)

Because the journalist claims that the proportion of very happy U.S. adults is a minority, that is, less than 0.50, the alternative hypothesis is:

\(H_A \colon p < 0.50\)

The test statistic is therefore:

\(Z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.47-0.50}{\sqrt{\dfrac{0.50(0.50)}{1052}}}=-1.946\)

Now, this time, we need to put our critical region in the left tail of the normal distribution. If we use a significance level of \(\alpha = 0.05\), then the critical region is:

Zα = 0.05-1.645

That is, we reject the null hypothesis if the test statistic \(Z < −1.645\). Because the test statistic falls in the critical region, that is, because \(Z = −1.946 < −1.645\), we can reject the null hypothesis in favor of the alternative hypothesis. There is sufficient evidence at the \(\alpha = 0.05\) level to conclude that \(p < 0.50\), that is, U.S. adults who are very happy are in the minority. The journalist's pessimism appears to be indeed warranted.

Note that this is an example of a left-tailed hypothesis test because the action falls in the left tail of the normal distribution.